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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 4, Problem 4.7.75

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 3x⁻²ᐟ³, y(−1) = −5

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Identify the given differential equation and initial condition: \( \frac{dy}{dx} = 3x^{-\frac{2}{3}} \) with \( y(-1) = -5 \).
Rewrite the differential equation in integral form: \( dy = 3x^{-\frac{2}{3}} dx \).
Integrate both sides with respect to \( x \): \( y = \int 3x^{-\frac{2}{3}} dx + C \), where \( C \) is the constant of integration.
Use the power rule for integration: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \), making sure to add 1 to the exponent \( -\frac{2}{3} + 1 = \frac{1}{3} \).
Apply the initial condition \( y(-1) = -5 \) to solve for \( C \) by substituting \( x = -1 \) and \( y = -5 \) into the integrated expression.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing the variables to be separated on opposite sides of the equation. This technique simplifies solving by integrating each side independently.
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Integration of Power Functions

Integrating power functions involves applying the power rule for integration, which states ∫x^n dx = (x^(n+1))/(n+1) + C, for n ≠ -1. Understanding how to handle fractional and negative exponents is essential for solving the given differential equation.
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Initial Conditions and Particular Solutions

Initial conditions specify the value of the solution at a particular point, allowing determination of the constant of integration. This transforms the general solution of a differential equation into a unique particular solution that satisfies the given condition.
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